Unit 9 Planar Graphs

Planar Graphs UNIT 9 PLANAR GRAPHS

Structure Page No

9.1 Introduction 65 Objectives 9.2 Embeddings and Planarity 66 9.3 Dual Graphs 68 9.4 Euler’s Formula 70 9.5 Colouring of Planar Graphs 73 9.6 Summary 75 9.7 Hints/Solutions 75 9.8 Appendix-Four Colour Problem 78

9.1 INTRODUCTION

The study of graph layouts and embedding of graphs on surfaces is termed as topological graph theory. Topology Graph Theory was first described by Euler in 1736 and later developed by Kurotowski, Whitney and others.

In earlier days the study of the concept of planarity was motivated by famous problems such as four colour problem and the three utility problems, etc. Now a days this concept is also applied in the design of circuit layouts on silicon chips!

A graph whose diagram can be drawn on a surface, on the plane or on the surface of a sphere, without over-crossing of edges is called a planar graph and such a drawing is called planar representation of the graph. In this unit we will investigate characteristics of planar graphs. We will study concept of dual graphs and colouring of planar graphs. Recall that the colouring of graphs is discussed in the previous unit, i.e. Unit 8. We shall also discuss here the famous four colour problem and will give a proof of the 5 colour theorem. A brief history of the four colour problem is also given as an Appendix.

Objectives

After studying this unit, you should be able to

• identify planar graphs and non planar graphs;

• prove that K 5 and K 3,3 are not planar;

• identify the faces of a plane graph;

• draw the dual of a plane graph;

• apply Euler’s formula for plane graphs;

• explain four colour problem for planar graphs.

Graph Theory 9.2 EMBEDDINGS AND PLANARITY

In this section, we study planar embeddings of planar graphs and two important graphs which play key roles in the characterization of planar graphs.

You can start reading the Text Book DW.

READ DW page 233 line 1 to page 235 line 21

NOTES:

(i) Page 233 line 1

Topological graph theory deals with embeddings of graphs on surfaces which resulted in the theory of planar graphs.

(ii) Page 233 line 2

The four colour problem was about the minimum number of colours necessary for the coloring of the regions in a map with no pair of region sharing boundary receives the same colour.

(iii) Page 233 line 8

The gas-water-electricity problem discussed in Example 6.1.1 of Textbook DW is called the three utility problems. The question is whether it is possible to lay pipes/cables from the source of each of the three utilities to three different consumers without over-crossing

the lines. The answer is affirmative since its graph model is K 3,3 , which cannot be drawn on a plain without over-crossings. You may try various drawings and convince the fact yourself intuitively. You will see a formal proof later in proposition 6.1.2 page 234 of

Text book DW. A drawing of K 3,3 is given in page 233 and another on page 234 of Textbook DW.

(iv) Page 234 line 5

K 5 and K 3,3 are the two graphs responsible for non-planarity of graphs. They and their subdivisions forbid a graph to be planar. A

geometrical proof of non-planarity of K 5 and K 3,3 is given in 6.1.2.

First draw a spanning cycle (cycle containing all the vertices, which exists both the cases) in the graph then try to draw the remaining edges. It can be seen that some of the remaining edges cannot be drawn without over-crossing. An analytical proof is given in Example 6.1.24 as a consequence of Theorem 6.1.23 of Textbook DW.

(v) Page 234 line 19 to page 235 line 5 Planar Graphs

A planar graph can be drawn on the plane without over-crossing of edges. Over-crossing means the edges have common point in the plane other than their end vertices. Such a drawing is called planar embedding (some authors call plane embedding) of the graph.

A planar graph may have more than one planar embeddings.

For example, Fig. 1 given below contains three drawings of the

planar graph K 4 , but only the second and the third are planar graphs.

(a) (b) (c) Fig. 1

(vi) Page 235 line 5 to page 235 line 21

Concepts of open set, region and faces are intuitive. Open set is similar to that in the real plane. Region is a connected open set and face is interior of cycle (or possibly closed walk in case of cut edge) not containing vertices or edges not belonging to the cycle. A cycle has two faces, inner (bounded) and outer (unbounded) and a tree has only one face. Fig.2

Fig. 2

Consider the graph in Fig. 2. It is a planar embedding of a graph with four vertices u, v, w, x and four edges a, b, c, d. There are two faces for this graph viz. the inner and outer. The cycle uvw is the boundary of the inner face and the boundary of the outer face is the walk ubwdxdwcvau, which is not a cycle. If one draws the vertex x inside the cycle u v w, then the edge d and the vertex x will be on the boundary of the inner face. Whenever we count the faces of a planar embedding, the outer surface also need to be counted.

Now we want you to try some exercises.

E1) Show that every subgraph of a planar graph is planar.

E2) Is every subgraph of a non-planar graph non-planar? Justify. 67

Graph Theory E3) Show that the graph given below is planar.

Fig. 3

E4) Determine all values of n such that the complete graph Kn is planar.

In the next section we shall consider the ‘dual’ of a graph.

9.3 DUAL GRAPHS

Duality is an important and interesting concept in the study of planar graphs. Let us try to understand this concept. A plane graph G partitions the rest of the plane into a number of connected regions; the closures of these regions are called the faces of G. Fig. 4 shows a planar graph with six faces, f1, f2, f3, f4, f5, and f6. The notion of a face applies also to embeddings of graphs on other surfaces. We shall denote by F (G) and φ(G), respectively, the set of faces and the number of faces of a plane graph G.

Fig. 4

Each plane graph has exactly one unbounded face, called the exterior face; in the plane graph of Fig. 4, f is the unbounded face (i.e. the face which is not shaded).

Let us try to understand the concept of duality through an example. We shall consider the planar embedding G of the cube as given in Fig. 5 (a).

(a) (b) (c) Fig. 5

Now we place a new vertex within each face, (including the unbounded face) Planar Graphs and join the pairs of new vertices in adjacent faces. Then we obtain the graph G*, which is the planar embedding of an octahedron (see Fig. 5. (c). Then G* is called the dual of G. The new vertices are represented by small circles, and lines joining them are indicated by dash lines. [See Fig.5 (b) and 5 (c)].

To understand the formal definition, you can start reading Textbook D.W.

READ Textbook DW from page 236 line 16 to 240 line 22

NOTES: i) Page 236 Line 16 to page 237 line 26

A planar embedding of K 4 and its dual are given by bold and dashed

edges respectively in Fig.6. Note that the dual of K 4 is itself.

Fig. 6

The dual of a planar graph depends on the planar embedding and hence need not be unique.

In a dual, leaves will appear as loops and vertices of degree two as multiple edges.

The dual of the graph in Fig. 2 is the graph with two vertices, triple edge joining them and a loop at one of the vertices. The dual of a tree with e edges is just a vertex and e loops at it. The dual of a cycle of length n is simply two vertices and n edges joining them. ii) Page 237 Line 27 to page 238 line 22

Theorem 6.1.12 in Textbook DW states that irrespective of the drawing, the sum of lengths of faces of a planar embedding is twice the number of edges. The counting method discussed above is justified by this result.

While counting the edge on the boundary of a face, cut edges are to be counted twice as both sides of it represent same face, probably outer. If there is a leaf in a planar graph, its planar embedding can be drawn with 69

Graph Theory either as the leaf is on the boundary of the outer face or on the boundary of an inner face.

iii) Page 239 Line 9 to page 239 line 24

Theorem 6.1.16 of Textbook DW gives two characterisations for the dual of a planar graph to be Eulerian.

iv) Page 239 Line 25 to page 240 last line

A particular class of planar graphs, outer planar graphs, is discussed here. They are planar graphs whose planar embeddings can be drawn in such a way that all vertices lie on the boundary of the outer face.

Read the proposition 6.1.18 in Textbook DW as “the boundary of the outer face of a 2-connected outer plane graph is a spanning cycle”.

The proposition 6.1.19 in Textbook DW gives two planar graphs which are not outer planar and the proposition 6.1.20 gives a necessary condition for outer planarity, which is helpful in checking outer planarity of a simple graph.

Here are some exercises for you.

E5) The following diagrams show two different plane drawing of the same planar graph. Show that their duals are not isomorphic.

(a) (b) Fig.7

E6) Which of the following statements are true? Justify your answer.

i) If a planar graph has a cut vertex, then its dual has a cut vertex.

ii) If the dual of the planar graph has a cut vertex, then the graph itself has a cut vertex.

In the next section we shall consider a remarkable simple formula that relates the number of edges, vertices and faces of a planar graph.

9.4 EULER’S FORMULA

Here we are going to discuss the formal proof of a formula that has been taught in school classes associated with the study of solids. It is the formula 70 relating the vertices (n), edges (e) and faces (f) in planar graph and

the formula is − + fen = 2 . It is easy to see that a stereographic projection of Planar Graphs a solid on plane gives a planar graph with vertices of the solid as vertices of the graph and edges of the solid as edges. To learn more about this, you can start reading the Textbook DW.

READ Textbook DW chapter 6 pages 240 to 243

NOTES: i) Page 241 Line 1 line 16

We give an alternative proof of Theorem 6.1.21 on page 241 of Textbook D.W.

Proof: we use induction on e, the number of edges. Basic Step: e = 0

G is connected and so = KG 1

∴n =1, and f = 1 ∴ − + = 2fen

Induction step (e>1): Now let G be a graph as given in theorem and assume that the theorem is true for all connected plane graphs with less than e edges.

Case 1: If G is a tree, then n = e + 1 and f = 1 So − + = + − eefen + = 211

Case 2: If G is not a tree, then G contains a cycle. Let x be an edge of some cycle in G and consider ' −= xGG . Since x being an edge on a cycle, it is in the boundary of two faces of G and hence its deletion merges the two faces. So G ' has n vertices, −1e edges and f − 1 faces only. Therefore by induction hypothesis, G ' satisfies the Euler’s formula i.e. − − + − = 2)1f()1e(n and consequently − + = 2fen . Hence the theorem is proved. □□□ ii) Page 241 Line 17 to page 242 line 24

It has been noted that the dual of planar graph depends on the planar representation. But the total number of vertices, edges and faces in the dual remain constant irrespective of the planar representation.

The Euler’s formula given in 6.1.21 is only for connected graphs.

Graph Theory For a disconnected graph the components share the outer face and hence the contribution by the number of components is also to be incorporated. So for a disconnected graph with k components the Euler’s formula generalizes as n - e + v = k + 1. In case of connected graph =1 and it reduces to − + ven = 2

Here we give proof of the generalized formula which can be stated as follow.

Theorem: 1 Let G be a plane graph with n vertices, e edges, f faces and k components. Then − + = + 1kfen .

Proof: Consider a plane embedding of G such that the exterior face is L shared by all components. Let the components be 21 ,,, GGG k .

L Let G i have n i vertices e i edges and f i faces for = ,,2,1 ki .

Then for each i = 1 to k we have iii =+− 2fen by Euler’s formula. k k k LLL Summing over i, i i ∑∑∑ i =+− 2kfen )1( i=1 i=1 i=1

k k

But, ∑ i = nn , ∑ i = ee and i=1 i=1 k

∑ i −+= (ff )1k since the exterior face is counted once for each of the =1i k

k components in ∑ f i and it counts only once in f. =1i So (1) becomes − + + − = 2)1( kkfen

Thus − + = kfen +1 □□□

iii) Page 241 Line 27 to Line 28

Theorem 6.1.23 gives necessary conditions that can be easily checked. The first is for general graphs while second condition is much stronger

only for triangle free graphs. A triangle is a K3 subgraph.

iv) Page 242 Line 3 to Line 5

Example 6.1.24 gives an analytical proof of non-planarity of K 5 and

K ,3 3 using the necessary conditions described in 6.1.23.

We shall now use Euler’s theorem to provide another interesting result.

Proposition 1: If G is a connected planar simple graph with at least 3 vertices Planar Graphs 5 and e edges whose shortest cycle length is 5, then ()ne −≤ 2 . 3 Proof: We first note that the shortest cycle length in G is 5. Therefore the 2 degree of each face in a plane drawing is at least 5, so that ≥ 52 fe or ≤ ef . 5 Combining this with Euler’s formula, = − nef + 2 we get, 2 2 ≤+− ene 5 2 nee −≤− 2 5 3 ne −≤ 2 2 5 ()ne −≤ .2 3 Hence the result.

□□□

Here are some exercises for you.

E7) Let us verify Euler’s formula for the following graphs

(a) (b) Fig.8

E8) Let G be a plane graph with n vertices, e edges and f faces. If every − )2n(k face of G is a k-cycle, then e = . k − 2 E9) Use proposition 1 to show that the Peterson graph is not planar.

In the next section, we discuss colouring of planar graphs.

9.5 COLORING OF PLANAR GRAPHS

Recall the concept of vertex coloring of graphs studied in Unit 8.

Coloring of planar graphs is related to coloring of a maps. A proper coloring of a map means an assignment of colors to the regions in the map such that no pair of regions sharing boundaries receive the same color. For every map it is possible to obtain a planar graph whose vertices representing the regions and two vertices are joined by an edge if and only if the respective region share

Graph Theory boundary. Then the map coloring problem becomes the vertex coloring problem in planar graphs.

There was a classical problem called the Four Colour Problem paused in 1852 by Francis Guthrie, a student of geography, about the minimum number of colours necessary for a proper colouring of a map. The problem was communicated to De Morgan through his brother Frederic Guthrie. Cayley published a paper on this problem in 1879 and outlined the difficulties that lie in obtaining a solution. In the same year A.B. Kempe came forward with a proof. After ten years, in 1890, P.J. Heawood detected an error in the proof and he modified it and proved the sufficiency of five colours − the five colour theorem. Finally, after about a century, in 1979 Appel and Haken put an end by proving the sufficiency of four colours and hence now we have the four colour theorem.

You will learn more about this while reading the Textbook D.W.

READ DW Chapter, Section, from Page 257 - 258

NOTES: i) Page 257 Line 7

Here we shall prove the following theorem.

Theorem:2 Every simple connected planar graph with n vertices has at most − 36n edges. (Hence such a graph has a vertex of degree at most 5).

Proof: ≤ − 6n3e (by Theorem 6.1.23)

∴ ≤ −12n6e2

∑ −≤ 12n6vdegv where deg v is the degree of any vertex v. v

Since G is connected, deg ≥1v ∀ vertices v. If all the vertices have deg ≥ 6 , then ∑ ≥ n6vdeg . v But ∑ −≤ 12n6vdeg v

Hence there exists a vertex of degree at most 5. In fact we can say that at least three vertices have degree less than 6.

Here are some exercises for you to try. Planar Graphs

E10) Which of the following statements are true? Justify your answer i) The chromatic number of a planar graph cannot exceed 4. ii) Every planar graph is not 6-colourable.

E11) Check whether the Four-Colour problem holds for the following graph.

Fig.9

9.6 SUMMARY

In this unit, we have covered the following points:

1. The concepts of planarity, planar embedding, outer planarity, faces, dual of a plane graph.

2. Non planarity of K 5 and K 3,3

l 3. If i )F( denotes the length of face Fi in a plane graph G, then l = ∑ FGe i )()(2 i

4. Edges in a plane graph G form a cycle in G if and only if the corresponding dual edges form a bond in G*.

5. K 4 and K 3,2 are planar but not outer planar.

6. The Euler’s formula.

7. Necessary relationship between the order and size of a planar graph in the general situation and in triangle free graphs.

8. Every planar graph is 5 – colourable.

9.7 HINTS/SOLUTIONS

E1) Let G be a planar graph. Let H be a subgraph of G. Then ⊆ )G(V)H(V and ⊆ E)H(E )G( . If G can be drawn on a plane without crossings then H also can be drawn on a plane without crossings. Hence H is planar.

E2) K 5 is non-planar but 5 }e{\K is planar where e is any edge of K 5 .

Graph Theory E3) The given graph can be drawn as given below two edges meet except at a vertex with which they are both incident.

E4) Since K4 is planar and Kn is a subgraph of K4 when n ≤ 4, Kn is planar when n ≤ 4. Since K5 is non-planar, and K5 is a subgraph of Kn when n ≤ 5 , Kn is non-planar when n ≤ 5 .

E5) The dual graphs are given respectively in Fig, 10 (a) and (b).

(a) (b) Fig.10

Their degree sequences are (3, 3,3,3,3 ,5) and (3,3,3,3,4,4) and therefore they are not is isomorphic.

E6) i) If a plane graph G has a cut vertex then if its dual has a cut vertex.

This statement is not true. The following figure illustrates this.

(a) (b) Fig.11

In Fig.11 G has a cut vertex v. But its dual G* has no cut vertex.

ii) This is not true as the following example shows.

(a) (b) Fig.12

In Fig.12 G* has a cut vertex v. But G does not have a cut vertex.

E7) i) Here n= 10, e = 15 and f = 7. ii) Hint: Direct Verification. 76

E8) Every face of G is a k-cycle. Hence every edge lies on the boundary of Planar Graphs exactly two faces.

∴ = k.fe2 e2 f =∴ k By Euler’s formula, − + = 2fen e2 i.e. en =+− 2 k  2  i.e. e −  −= n21  k 

− )n2(k e =∴ 2 − k

− )2n(k i.e. e = k − 2

E9) Suppose that the Peterson graph is planar. Then the inequality in 1 example 2 becomes ≤ 1315 , which is not possible. Hence the peterson 3 graph is not planar.

E10) i) True –since any planar graph is plane graph. ii) False-follows from (i) or it can be shown directly that every planar graph is G-colourable.

E11) We first label the vertices a,…,f as follows:

Fig.13

Then we successively colour the vertices as Vertex a with colour green Vertex f with colour blue Vertex c with colour green Vertex d with colour blue Vertex e with colour pink Vertex f with colour yellow

Fig.14

All the vertices are now coloured. We thus obtain the 4-colouring of G as show in Fig.14. 77

Graph Theory 9.8 APPENDIX − A BRIEF HISTORY OF THE FOUR COLOUR PROBLEM

The subject was first raised in 1852 by a part-time mathematician, Francis Guthrie, who was colouring in a map of the counties of Britain. He was intrigued by the fact that four colours appeared to be sufficient regardless of the complexity of boundary shapes or how many regions had a common border. He passed the problem on to University College London to the eminent De Morgan, who in turn passed it to the great William Hamilton. Hamilton was unable to invent a map which required five colours, but neither could he prove that no such map existed.

Like Fermat's Last Theorem this apparently trivial problem generated a great deal of interest and activity. In 1879 a British mathematician, Alfred Kempe, published a 'proof' that was accepted by the mathematics establishment until in 1890 Percy Heawood of Durham University showed that the so-called proof was fundamentally flawed. The search continued, and like Fermat's theorem led to great advances in number theory, so the four-colour problem gave a stimulus to the new and increasingly important topic of topology.

The first breakthrough in the four-colour problem came in 1922, when Philip Franklin ignored the General problem and settled for a proof which showed that any map containing 25 or fewer regions required only four colours. This was extended in 1926 to 27 regions, in 1940 to 35 regions, and in 1970 to 39 regions. Then in 1976 two mathematicians at the University of Illinois, Haken and Appel, came up with a new technique which would revolutionise the concept of mathematical proof.

Haken and Appel used the ideas of Heinrich Heesch that the infinity of infinitely variable maps could be constructed from a finite number of finite maps. They reasoned that by studying these building-block maps it would be possible to attack the general problem.

This proved very difficult in practice to achieve, because the number of building block configurations could not be reduced below 1482. To crank through all the permutations that might occur with this number of configurations would take a lifetime. Enter the age of the computer. In 1975, after five years of working on the problem, they turned to the new number- cruncher and in 1976 after 1200 hours of computer time they were able to announce that all 1482 maps had been analysed and none of them required more than four colours.

AND THUS WE HAVE THE LONG AWAITED FOUR COLOUR THEOREM.

The problem with this type of proof is that only another computer can carry out the customary check on its validity. Some mathematicians are most reluctant to accept it because no one, however patient, could work through the exposition line by line and verify that it is correct. It has been disparagingly referred to as a 'silicon proof'. The fact is, however, that mathematicians will increasingly have to rely on such methods. The age of the purist of pure logic as the only acceptable technique in mathematics has probably passed. 78